Answer:Best corral has a width and length of 50 feet, enclosing an area of 2500 square feet. Let's calculate the width (W) of the rectangle as a function of it's length (L). So we have W = (200 - 2*L)/2 W = 100 - L Now the area of the rectangle is A = WL Substitute the equation for W. A = (100 - L)L A = 100L -L^2 Let's make an initial guess of 40 ft and add an error component of e. So we'll use the length of (40+e) and see what we get. A = 100L -L^2 A = 100(40+e) - (40+e)^2 A = 4000 + 100e - (1600 + 80e + e^2) A = 4000 + 100e - 1600 - 80e - e^2 A = 2400 + 20e - e^2 Now looking at those 2 "e" terms is interesting. It's pretty obvious that any negative value of e will cause those term to result in a value less than 0, and decrease the available area. Also any value of e greater than 20 will also cause those 2 values to sum to a negative value and decrease the area. But a value of e in the range of 0 to 20 will result in a positive value and cause the area enclosed to be larger. So it's obvious that 40 feet isn't optimal. Let's pick the middle of the e values that result in something positive (0+20)/2 = 10 and add that to our initial guess, getting a length of 50 and replace length by (50+e) and see what happens. A = 100L -L^2 A = 100(50+e) -(50+e)^2 A = 5000+100e -(2500+100e + e^2) A = 5000+100e - 2500 - 100e - e^2 A = 2500 - e^2 This looks quite promising. Any non-zero value of e will result in the area enclosed being smaller. So the idea value of e is 0. That means that the idea length of the rectangle is 50 feet. And that makes the width 50 feet as well. Mind, this problem could have been also solved using the first derivative of the equation A = 100L -L^2, which would have been A' = 100 - 2L, and then solving for 0. But I did this problem to demonstrate that you don't need to resort to calculus for every maximum type of problem.
